Proof of convergence of Dirichlet's Eta Function I'd like to check directly the convergence of Dirichlet's Eta Function, also known as the Alternating Zeta Function or even Alternating Euler's Zeta Function:
$$\eta(s) = \sum_{n=1}^\infty\frac{(-1)^n}{n^s}\;\;,\;\;\;s=\sigma+it\;,\;\;\sigma\,,\,t\in\Bbb R\;,\;\;\color{red}{\sigma > 0}.$$
Now, there seems to be a complete absence of any direct proof of this in the web (at least I didn't find it) that doesn't use the theory of general Dirichlet Series and things like that.
I was thinking of the following direct, more elementary approach:
$$n^{it}=e^{it\log n}:=\cos (t\log n)+i\sin(t\log n)$$
and then we can write
$$\frac1{n^s}=\frac1{n^\sigma n^{it}}=\frac{\cos(t\log n)-i\sin(t\log n)}{n^\sigma}$$
and since a complex sequence converges iff its real and imaginary parts converge, we're left with the real series
$$\sum_{n=1}^\infty\frac{\cos(t\log n)}{n^\sigma}\;\;,\;\;\;\;\sum_{n=1}^\infty\frac{\sin(t\log n)}{n^\sigma}$$
Now, I think it is enough to prove only one of the above two series' convergence, since for example $\;\sin(t\log n)=\cos\left(\frac\pi2-t\log n\right)\;$
...and here I am stuck. It seems obvious both series are alternating but not necessarily
elementwise.
For example, if $\;t=1\;$ , then  $\;\cos\log n>0\;,\;for\;\;n=1,2,3,4\;$ , and then $\;\cos\log n<0\;,\;\;for\;\;\;n=5,6,\ldots,23\;$ . This behaviour confuses me,  and any help will be much appreciated.
 A: I think this succumbs to applications of Dirichlet's test and some estimates based on the mean value theorem.
Dirichlet's test gives conditional convergence of a sum $\sum_{n \geq 1} a_n b_n$ provided that $a_n$ is a sequence of positive numbers that decrease to $0$ and the sequence of partial sums $B_n = \sum_{k=1}^n b_k$ is uniformly bounded in absolute value. In this case, take $a_n = n^{-\sigma}$ and $b_n = (-1)^{n+1} n^{-it}$. Clearly the $a_n$ decrease to $0$ if $\sigma > 0$, so we just need to get the uniform boundedness of the $B_n$.
It suffices to bound $\sum_{k=1}^n (2k-1)^{-it} - (2k)^{-it}$. Clearly this is bounded for $t=0$, so assume $t\neq 0$. The real and imaginary parts are
$$C_n = \sum_{k=1}^n \cos(t\log(2k-1)) - \cos(t\log(2k)),$$
$$S_n = \sum_{k=1}^n \sin(t\log(2k)) - \sin(t\log(2k-1)).$$
Let's show how to bound the imaginary part $S_n$; the real part is bounded by a wholly analogous technique.
By the mean value theorem, the $k$-th term of $S_n$ is
$$\frac{t\cos(t\log(x))}{x}$$
for some $x \in [2k-1, 2k]$. This is not much different from $\frac{t\cos(t\log(2k-1))}{2k-1}$; in fact, the difference between them is, again by the mean value theorem,
$$-\frac{t^2 \sin(t\log(y)) + t\cos(t\log(y))} {y^2} (x - (2k - 1))$$
for some $y \in [2k-1, x]$, whose absolute value is bounded above by $\frac{t^2+t}{(2k-1)^2}$. Since $\sum_{k\geq 1} \frac1{(2k-1)^2}$ converges, this reduces us to putting a uniform bound on the absolute value of
$$\sum_{k=1}^n \frac{\cos(t\log(2k-1))}{2k-1} = \text{Re}\sum_{k=1}^n \frac1{(2k-1)^{1 + it}}.$$
Put $z = 1 + it$; we may as well bound the absolute value of
$$\sum_{k=1}^n \frac1{(2k-1)^z} = \sum_{k=1}^n \left[\int_k^{k+1} \frac1{(2k-1)^z} - \frac1{(2x-1)^z} dx\right] + \int_1^{n+1} \frac1{(2x-1)^z} dx.$$
The last integral is easily evaluated as $\left[\frac{(2x-1)^{-it}}{-2it}\right]_1^{n+1}$ which in absolute value is bounded above by $\frac1{t}$. We turn now to the preceding sum of integrals. The integrands are bounded thus:
$$\left|(2k-1)^{-z} - (2x-1)^{-z}\right| = \left|\int_{2k-1}^{2x-1} \frac{z}{t^{z+1}}\; dt\right| \leq \frac{2|z|}{(2k-1)^{\text{Re}(z) + 1}}.$$
Thus the sum of the integrals is bounded by the finite quantity $2|1+it| \sum_{k=1}^\infty \frac1{(2k-1)^2}$, and we are done.
